Question

Giúp em với ạ

Answer 1

Bài giải

  

Answer 2

Áp dụng BĐT Cô-si:

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=\dfrac{2a}{c}\)

Tương tự: \(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\ge\dfrac{2c}{b}\) ; \(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2b}{a}\)

Cộng vế:

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\right)\)

\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\)

Dấu "=" xảy ra khi \(a=b=c\)